The equation of the line that goes through the point (2,3) and is parallel to the 4x+3y=3 can be written in the form y=mx+b where m is: and b is:
+37146 Rewrite 4x+3y=3
3y=-4x+3
y = -4/3x +1 where =-4/3 is the slope....parallel line will have same slope
y = -4/3 x + b includes 2,3
3 = -4/3 (2) + b
3+8/3 = b
b = 5 2/3
so NEW line is y = -4/3x + 5 2/3 m = -4/3 b = 5 2/3
+1252 We can make \(4x+3y=3\) into \(y=mx+b\) form! We need to isolate \(y\) .
\(3y=-4x+3, y=-\frac{4}{3}x+1\)
However, the line goes through (2, 3) so the y-intercept may not be 1! Let's substitute.
\(3=-\frac{4}{3}\times 2+b, 3=-\frac{8}{3}+b, b=17/3\) and m=-4/3.
You are very welcome!
:P
